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SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

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Question : 131 of 200

Marks: +1, -0

Determine the current

i_a

in the given network

- 1 A
1 A
- 4 A
4 A

Solution:

Here,

i_{a}

is given by,

i_{a}=\frac{V_x-12}{4}

...(1)

Apply

\text{KVL}

at node

V_x

,

\frac{V_x-12}{4}+3i_a+\frac{V_x}{2}=0

Put the value of

i_a

in above equation,

\frac{V_x-12}{4}+3\left(\frac{V_x-12}{4}\right)+\frac{V_x}{2}=0

\frac{V_x-12}{4}+\frac{3V_x-36}{4}+\frac{V_x}{2}=0

V_x-12+3V_x-36+2V_x=0

6V_x=48

V_x=8\,\text{V}

Put this value of V in equation (1)

i_a=\frac{8-12}{4}=-1\,\text{A}

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